1

### JEE Main 2017 (Online) 8th April Morning Slot

Among the following, the incorrect statement is :
A
At low pressure, real gases show ideal behaviour.
B
At very low temperature, real gases show ideal behaviour.
C
At very large volume, real gases show ideal behaviour.
D
At Boyle’s temperature, real gases show ideal behaviour.

## Explanation

A real gas do not show ideal behaviour at low temperature, high pressure and low volume.

So according to the question, at low temperature real gas show ideal behaviour. This statement is wrong.
2

### JEE Main 2017 (Online) 9th April Morning Slot

At 300 K, the density of a certain gaseous molecule at 2 bar is double to that of dinitrogen (N2 ) at 4 bar. The molar mass of gaseous molecule is :
A
28 g mol$-$1
B
56 g mol$-$1
C
112 g mol$-$1
D
224 g mol$-$1

## Explanation

Density = ${{Mass} \over {Volume}}$

PV = RT $\Rightarrow$ V = ${{RT} \over P}$

So, Density(d) = ${{MP} \over {RT}}$

Now, d1 = x, P1 = 4, M1 = 28, d2 = 2x, P2 = 2, M2 = ?

$\therefore$ ${{{d_1}} \over {{d_2}}} = {{{M_1}{P_1}} \over {R{T_1}}} \times {{R{T_2}} \over {{M_2}{P_2}}} = {{{M_1}{P_1}} \over {{M_2}{P_2}}}$ [As T1 = T2 ]

$\Rightarrow$ M2 = ${{{M_1}{P_1}{d_2}} \over {{d_1}{P_2}}}$

= ${{2x \times 28 \times 4} \over {2 \times x}}$ = 112 g mol-1
3

### JEE Main 2018 (Online) 16th April Morning Slot

Assuming ideal gas behaviour, the ratio of density of ammonia to that of hydrogen chloride at same temperature and pressure is : (Atomic wt. of Cl = 35.5 u)
A
1.46
B
0.46
C
1.64
D
0.64

## Explanation

We know, PV = nRT

n = no. of moles = ${m \over M}$

So,   PV = ${m \over M}RT$

$\Rightarrow $$\,\,\,\, P = {m \over V} \times {{RT} \over M} \Rightarrow$$\,\,\,\,$ P = d $\times$ ${{RT} \over M}$ [ d = density = ${m \over V}$ ]

at constant temperature and pressure d $\propto$ M

Now let d1 and d2 are the density of ammonia and HCl.

$\therefore\,\,\,\,$ ${{{d_1}} \over {{d_2}}} = {{{M_{N{H_3}}}} \over {{M_{HCl}}}}$

$\Rightarrow$ $\,\,\,\,$ ${{{d_1}} \over {{d_2}}}$ = ${{17} \over {36.5}}$ = 0.46
4

### JEE Main 2019 (Online) 9th January Morning Slot

0.5 moles of gas A and x moles of gas B exert a pressure of 200 Pa in a container of volume 10 m3 at 1000 K. Given R is the gas constant in JK$-$1mol$-$1, x is :
A
${{2R} \over {4 + R}}$
B
${{2R} \over {4 - R}}$
C
${{4 + R} \over {2R}}$
D
${{4 - R} \over {2R}}$

## Explanation

We know,

PV = nRT

Given,

P = 200 Pa

V = 10 m3

T = 1000 K

n = 0.5 + x

$\therefore$   200 $\times$ 10 = (0.5 + x) R $\times$ 1000

$\Rightarrow$    0.5 + x = ${2 \over R}$

$\Rightarrow$    x = ${2 \over R} - {1 \over 2}$

$\Rightarrow$   x = ${{4 - R} \over {2R}}$